# Relativistic particle decay

## Homework Statement

A particle with momentum p0, mass m0 and energy E0 decays into two particles with mass m1 and m2. Find the energy of the particle E1 and E2.

Four-momentum!

## The Attempt at a Solution

I calculated the energy of particle 1 in S' (system where particle 0 is at rest) in dependence of m0', m1', m2'. I got also the momentum of the particle 1 in system S' which I can write in dependence of m0', m1', m2' and E1'. For example I have taken the coordinates of system S' in such a way, that I can write the momentum p'1 of particle 1 as: p'1=(0,py',0) with py'=sqrt(f(m0',...)). Thus getting the four-momentum of particle 1: (E', 0, c*py', 0)

Well, what my question concerns. How can I translate the whole thing back into system S? Lorentz-Transformation or is there a easyer way? How can I go on to fullfill the task?

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Dick
Homework Helper
The problem can't be solved as stated. E1 and E2 aren't fixed by knowledge of the initial energy and masses alone.

Well, the diffraction angle of p1 relative to p0, the energy E0 and momentum p0 are given as well as the masses m0,m1,m2.

Dick
Homework Helper
So you have an angle between p0 and p1, right? That's different... Then write the four momentum equation (E0,p0)-(E1,p1)=(E2,p2) and square both sides (in the Lorentzian sense). What do you get?

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Ok, this way I can solve it easy. It is no problem. I want to try it the other way. I mean, I know the solution of the problem in the frame of the moving particle p0 and want to transform it into the lab system (my system).

Dick
Homework Helper
Ok, this way I can solve it easy. It is no problem. I want to try it the other way. I mean, I know the solution of the problem in the frame of the moving particle p0 and want to transform it into the lab system (my system).
Well then, take a specific solution in the rest frame, like p1 and p2 along the y-axis) and figure out what direction you have to boost in to get the required angle. Sounds masochistic to me though.

I have a question for solving a similar problem: A particle of mass M at rest decays into two particles of masses m1 and m2. Show that the totatl energy of the first particle is E1 = c(sqr) (M (sqr)+ m1 (sqr) - m(sqr) ) / (2M). and that E2 is obtained by interchanging m1 and m2.
Can you please give me a hint how can I start working it. I know I have to use the energy-momentum 4-vector but do I use only one component for P?

Dick
Homework Helper
Both outgoing particles have the equal and opposite momenta - so you can assume there is only one component if you like. You know E1+E2=M and E^2-p^2=m^2 for each particle. So that's three equations. Eliminate p and E2 to get your expression. You'll have to fill in the c's - I set them to 1.

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I am still confused with this problem, I have the following equations:
E1+E2=MC^2 .................................(1)
E1^2-(p/C)^2=m1^2(C^4)...............(2)
E2^2+(p/C)^2=m2^2(C^4).............. (3) [p is equal in magnitude but oppsite in direction]

E1^2 + E2^2 = ( m1^2 + m2^2) C^4

Plugging the value of E2 from 1 gives a quadratic eguation:

2 E1^2 - 2MC^2 E1 + M^2C^4 = (m1^2 +m2^2 )C^4

Solving this equation doesn't give me the needed answer! Am I doing something wrong?

Dick
Homework Helper
(-p/c)^2=(p/c)^2. The square of a negative is positive. You have a flipped sign.

Oh, I got it now! Thank you so so much!

I have the following problem: In a collision process a particle of mass m2, at rest in the lab frame is struck by a particle of mass m1 with energy E(L), and momentum P(L). In the centre of mass frame the two particles will have 3-momenta p, and -p respectively. Show that the Lorentz boost parametre B(CM) and describing the velocity of the centre of mass frame in the laboratory is: B(CM) = cP(L)/ (m2c^2+E(L))

I started my solution: B= V(CM)/C
P= [gamma](CM)*m2* V(CM)
E2= [gamma](cm)*m2*c^2
E1= [gamma](cm)*m1*c^2....... (not so useful)

now when trying to solve the first the two equations I get:

V(CM)= P*C^2/E2 , this means that E2 = m2c^2+ E(L)! Does this answer make sense? The energy of the second particle in CM frame is equal to total "spatial" energy in Lab frame! How can this be true?

Dick
Homework Helper
I don't think you are doing a full lorentz boost, it should look more like this:

$$\left(\begin{array}{ccc}\gamma&-\beta \gamma\\-\beta \gamma&\gamma\end{array}\right)\left(\begin{array}{c}E\\p\end{array}\right) = \left(\begin{array}{c}E'&p'\end{array}\right)$$

Apply this to (m2 0) and (EL PL) and set the resulting momenta to be negatives of each other and it's dead simple.